This is only an approach of the problem. We have $P_n = a^n u_n$ where $u_n = \prod_{k=1}^n |\sin k|$. Clearly, $u_n$ converges to $0$, hence $\log u_n$ tends to $-\infty$.
Proposition Let $\ell = \limsup \frac{\log |\log u_n|}{\log n}$. If $\ell > 1$ then no such $a$ exists.
Proof: if $\ell > 1$, let $1 <\alpha< \ell$. For an infinite number of $n$ we have $|\log u_n| \ge n^\alpha$, hence $\log u_n\le -n^\alpha$, hence $\log P_n\le n \log a - n^\alpha$ which tends to $-\infty$ for this subsequence of ns.
Now have you tried this $\limsup$ numerically?
Second question
Can we expect that for large $n$
\begin{equation}
\frac{1}{n}\sum_{k=1}^n \log|\sin k| \approx \frac{1}{\pi}\int_0^\pi\log|\sin\theta| d \theta\quad?
\end{equation}
This question is closely related to computing the discriminant of Chebyshev polynomials of the second kind, found in literature (e.g., see G. Szegő Orthogonal polynomials, theorem $6.71$).
I'm going an elementary way here. Denote $x_{n,k}=\cos(k\pi/n)$ for $0\leqslant k\leqslant n$, so that $$S_n(x)=2^{n-1}\prod_{k=0}^n(x-x_{n,k})$$ is our polynomial. Then "ignoring the $1$ in the $\sqrt{1+(\dots)^2}$" amounts to saying that $l_k:=l_{n,k}$ is approximately twice the supremum of $\big|S_n(x)\big|$ on $x_{n,k-1}<x<x_k$; more precisely, we have $$L=\lim_{n\to\infty}\prod_{k=1}^n l_{n,k}=\lim_{n\to\infty}\prod_{k=1}^n\color{LightGray}{\Big[}2\sup_{x_{n,k-1}<x<x_{n,k}}\big|S_n(x)\big|\color{LightGray}{\Big]}=\lim_{n\to\infty}2^n\prod_{k=1}^n \big|S_n(x_{n,k}')\big|,$$ where $x_{n,k}'$ (for $1\leqslant k\leqslant n$) are the roots of $S_n'(x)$.
From properties of resultants, we know that if $f$ is a polynomial of degree $d$, with leading coefficient $a$, roots $x_{d,k}$ for $1\leqslant k\leqslant d$, and roots of its derivative $x'_{d,k}$ for $0<k<d$, then $$\prod_{k=1}^{d-1}f(x_{d,k}')=\frac1{d^d a}\prod_{k=1}^d f'(x_{d,k}).$$
In our case $d=n+1$ and $a=2^{n-1}$, so that $$L=\lim_{n\to\infty}\frac{2L_n}{(n+1)^{n+1}},\qquad L_n=\prod_{k=0}^n\big|S_n'(x_{n,k})\big|.$$
From $S_n(\cos t)=-\sin t\sin nt$ we get $S_n'(\cos t)=n\cos nt+\cot t\sin nt$, hence $$S_n'(1)=2n,\quad S_n'(-1)=2n(-1)^n,\quad S_n'(x_{n,k})=n(-1)^k\quad(0<k<n)$$ and $L_n=4n^{n+1}$, thus $L=\lim\limits_{n\to\infty}8/(1+1/n)^{n+1}=8/e$ as claimed.
Best Answer
One might use a Riemann sum argument, leading to a known integral: $$\log C=-\int_0^1\log\left(\cos\frac{x\pi}2+\sin\frac{x\pi}2-1\right)\,dx.$$
Alternatively, use $C=\lim\limits_{n\to\infty}P_n^{-1/n}$ with $$P_n:=\prod_{k=1}^{n-1}\left(\cos\frac{k\pi}{2n}+\sin\frac{k\pi}{2n}-1\right)=2^{3(n-1)/2}\prod_{k=1}^{n-1}\sin^2\frac{k\pi}{4n}$$ to get an integral reducing to $\int_0^{\pi/4}\log\sin t\,dt$. Thus, using Catalan's constant $G$, $$C=e^{4G/\pi}\sqrt2\approx4.5395015097179854508819422150513221738\cdots$$
Similarly, via Euler–Maclaurin summation formula, one gets $$L=\lim_{n\to\infty}(C^n P_{n+1}/n)=4/C.$$